Question

If $\frac{a}{b}+\frac{b}{a}=1$ then $(a^3-b^3)$=?
(a) -3 (b) -2 (c) -1 (d) 0

Answer 1

$\frac{a}{b}+\frac{b}{a}=1$ , multiply both side by ab

$a^2+b^2=ab$

$a^2+b^2-ab=0-------\left(1\right)$

$a^3+b^3=(a+b)(a^2+b^2-ab)$ ,

put $a^2+b^2-ab=0$ from eq.(1)

$a^3+b^3=(a+b)\left(0\right)=0$,

correct option is (d) 0